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Why are these 3 planes so close?

I spotted these three planes flying east then turning south.  They were over the Hopi reservation and heading toward Winslow, AZ at 10:42 AM local time (MST), today February 5 2021.  They were at 15k to 20k feet or so.  Can anyone explain?

Perspective really messes with distance estimation, those planes are not very close to each other.

Welcome to aviation.SE! This question might be helpful.

It would also help address some of the theories that have been raised in comments and/or answers if you would specifically state that they stayed generally in the same position relative to each other, or at least that the distance from the first to the last stayed generally the same, the entire time that you had them in sight.

The primary reasons large airliner-sized aircraft would stay in formation is for a photo flight, or for reasons related to military operations.
Very few civilian airliner-sized aircraft are grey.
The two grey aircraft are surely military tankers, both of the same type.  The white aircraft is in position to receive fuel from the front aircraft.  Based on color and large size (relative to the grey aircraft), the white aircraft is a 747-based military aircraft, such as a VC-125 (the type used as Air Force One) or an E-4.  With this in mind, the size of the tankers suggests that they are KC-135's.  A close look at the grey aircraft shows that they appear to have engines located well outboard on the wings, i.e. have four engines on the wings rather than two, further supporting this identification.
It is very common for military tankers to fly in a loose "trail" formation like this.  I've seen it dozens of times.  Whenever I see airliner-appearing aircraft flying in this type of formation I always reach for my binoculars and they usually turn out to be KC-135's, KC-46's, or KC-10's.
As to the "why" part of your question, a possible answer might be that two tankers operating together gives both crews a chance to practice refueling another aircraft of some particular type-- in thus case the E-4 or VC-125-- without having to put two different receiver aircraft in the air.  There are many other reasons why the military often finds it advantageous to move several aircraft through the airspace together as a single formation rather than as completely independent flights.
FlightRadar24 is a good resource for answering a question like this-- it allows one to zoom in on a given area and play back tracks from a given time in the past.  Much military traffic is not captured, but one could at least verify the presence or absence of civilian traffic at the time and place of interest.  ASE user DeepSpace has now provided this link, showing that in this case the ADSB transponder of a KC-135R was in fact captured at the time and place of interest.  The flight started and ended in Phoenix, and included two passes over the Hopi reservation.  The reported altitude over the Hopi reservation at the time of interest was 22,075'.

Your assumptions are spot on! OP provided accurate date, time and location so using flightradar24 was the obvious next step. Lucky enough, the KC-135 was captured by ADSB https://i.imgur.com/3HXFDp1.png

@DeepSpace -- thanks, I wasn't aware that that resource would work so well, tried using FlightAware but it seems you need to know N-number, flight number, or origin or destination airport-- thanks for the lesson

Kudos quiet flyer, very nice!

The first two aircraft seem really close in lateral distance, but at much different altitudes. If they are separated by 4000-6000 feet of altitude (which could be the case), they are about a mile apart. That is a safe and respectable distance.
The third aircraft is about 30° (rough estimate) of displacement from the first two. Depending on the distance between the aircraft and the observer, this aircraft could b 1.25 miles from the first two. This is based on 2.5 miles of difference between the observer’s elevation and the aircraft’s altitude.

Valid points, but the chance of this NOT being an intentional formation flight is extremely slim.

Despite apparently this being an intentional formation +1, because 99% of the cases when two or more planes seem to be close, in fact they are not, far from it (pun intended).


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In the United States, are ultralights legally considered aircraft?
I sometimes hear it stated that in the United States, ultralight vehicles are not legally considered to be aircraft. For example, Wikipedia used to state:

In the United States ultralights are classified as vehicles and not aircraft, and are thus not required to be registered, nor is the pilot required to have a pilot certificate.

Under FAA regulations and US law, are ultralights considered to be aircraft?

Yes, under FAA regulations, ultralight vehicles are considered to be aircraft.
14 CFR 1.1 states:

Aircraft means a device that is used or intended to be used for flight in the air.

Ultralights are used or intended to be used for flight in the air, so they are aircraft.
14 CFR Part 103 defines the term "ultralight vehicle" and describes the rules for operating an ultralight vehicle. Although Part 103 avoids using the word "aircraft" to describe ultralight vehicles, nothing in Part 103 (or anywhere else) states that they are not aircraft.
It is true that ultralight vehicles are exempt from nearly all regulations governing aircraft. However, this isn't because ultralights aren't legally considered aircraft; it's because there are regulations which specifically make ultralights exempt from those regulations. For example:

Part 103, Section 7, states that

ultralight vehicles are not required to meet airworthiness standards,
pilots of ultralight vehicles are not required to have any kind of certificate, and
ultralight vehicles are not required to be registered.

Part 61, Section 51(j), states that pilots are only permitted to officially log flight time in aircraft that are registered (as well as military and public aircraft). Ultralights cannot be registered.
Part 91, Section 1(e), states that ultralight vehicles are exempt from Part 91 in its entirety. (If ultralight vehicles were not legally considered aircraft, then this paragraph would not need to exist!)

(Of course, a pilot may make logs of flights performed in ultralights, but the FAA does not recognize this time as "flight time".)
So, the FAA does consider ultralight vehicles to be aircraft. However, they are exempt from most—but not all—regulations which govern aircraft.

This is good information, but it's a bit absurd to assert that ultralight pilots may not log their time flying those types of aircraft. Sure, the FAA may not recognize such logged time, but flight logs are an important aspect of ultralight pilot skill proficiency programs and ratings. Also as a practical matter, one might have to produce logs when flying sites abroad (as I recently did in Germany.)

@MikeStrother Good point, I've edited the answer to try to clarify that.

I'm nitpicking the regulation, not your answer. :)

Got it, I'll edit the regulation to clarify that. :)

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text:
How does the angle of attack vary in turns?
I read somewhere that when the aircraft banks to turn, its angle of attack increases.Since angle of attack is the angle between the relative wind and chord line, how does turning increase it? Is it because the relative wind is now from different direction or is it because the wing that goes higher meets the air at a higher angle?

Given the variety of answers this is getting, I suggestyou try to clarify exactly what it is you're asking.

Added one more tag since question was at top of stack anyway due to another recent answer...

Just to avoid some confusion that I feel in your question:
It doesn't!
That is, it doesn't happen by itself. The pilot or autopilot must actively pitch up (and possibly do other adjustments, e.g. increase thrust). Without active control, the angle of attack will stay about the same (initially), and the airplane will start descending if it enters a turn, because the lift becomes insufficient for level flight (see pictures in other answers).

yes that was confusion - I thought AOA varies when aircraft enters into a turn. So I understand that it only varies when pilot or autopilot pitches it up to increase the lift

As soon as the aircraft descends, AoA will increase from the change in the flight path angle. No pitching needed, but now drag will increase and the aircraft both slows down and pitches down in order to stay at its trim point.

The aircraft turns to change its speed vector. The absolute value of the vector stays the same, but its direction is changed. A force is needed to change the direction of movement of any mass, and the wing is used to provide this force in addition to the lift force. This is why lift is increased in a turn.

For turning, the aircraft needs to add a force in the direction of the red vector in the sketch above. This it can either do by pointing its nose to the right, and then the fuselage will create a small lateral force, albeit at a heavy price in drag. Or the airplane uses its wing to provide this force, which is a much cleverer way of creating the desired force because the wing is by far the most efficient "force generator" of an aircraft.

Note that as the turn progresses, the red vector will point in the opposite direction of the initial (blue) speed vector, which means that the speed in the initial direction is reduced to zero when a 90° turn is completed. The horizontal component of the lift force accelerates the aircraft in the new, desired direction and decelerates it in the old direction.
Lift can be increased either by speeding up, by flying in denser air or by increasing the angle of attack. The first two options are not very practical, so it is the angle of attack change which gives the aircraft the additional lift for turning. This angle of attack increase is controlled by pulling the stick gently back, thereby reducing lift on the horizontal tail so the aircraft can pitch up.
If you need more formulas to calculate the precise angle of attack change, look at this answer for flight mechanics in banked flight and this answer for how lift depends on angle of attack.

can you explain what does this mean "speed in the initial direction is reduced to zero when a 90° turn is completed"

@user2927392: Yes. After turning 90°, all initial speed is gone. Instead, the aircraft has gained the same speed value in the perpendicular direction, and all the acceleration to reduce the initial speed and build up the new speed has been provided by the wing's lift. Think of speed as a vector here - the scalar quantity does not change in a turn.

The AoA in a turn will be higher than in level flight IF the same airspeed and level is maintained. Reason being, in a turn some of the "upwards" (relative to the lateral axis) force generated by the wing is used to turn the aircraft around, so more lift needs to be generated to maintain altitude. To generate more lift at the same airspeed, the angle of attack has to increase.
This image illustrates it nicely:

source: http://avstop.com/ac/flighttrainghandbook/forces.html
(consider the "Total Lift" vs. the "Vertical Component of Lift")

So if you do not pitch up during a turn, will the angle of attack remain same even though aircraft pitches down a bit during the turn

This does not explain how the AoA is increased.

@user2927392 Yes, you can maintain the same AoA, but you would have to sacrifice some altitude.

@Simon The question is not "how does the angle of attack increase?", it is "how does the angle of attack vary?"

"Since angle of attack is the angle between the relative wind and chord line, how does turning increase it." that's how I interpret it, and as a rotary pilot, it's not clear to me how the turn causes the angle of attack to increase so to a non-pilot, it most certainly will be confusing.  Perhaps you could edit to explain how - I assume back pressure on the yoke/stick to maintain altitude and thus increase the AoA.

Essentially it's a combination of the increased pitch needed to maintain level and the increased G-load caused by the centrifugal force. I tried to answer in a relatively simply way that most pilots will understand. I'll leave the more complex aerodynamics for someone else to explain :)

@user2927392 Do you feel like your question has been answered?  I seriously can't tell.

I believe as long as your only feeling 1g on the seat of your pants, then the AoA has not changed relative to what it was during straight and level at the same airspeed.  This doesn't take into account variations in the AoA at the points along the wing where the ailerons are as they may need to change their AoA to hold bank.

@RyanMortensen Intuitively I would agree, that for a constant airspeed and G-load, the AoA should remain the same. I don't have the formulas to support it, though.

@J.Hougaard Ditto. I'm basing my assumption on total lift generation being equal at the same G-force.

you may add the reference from where you found this picture

@RyanMortensen If you feel only 1g and the aircraft is banked then the aircraft is accelerating downward. If it was flying level before, it will not just descend, but descend at an increasing rate of vertical speed. To maintain a level turn, http://www.altairva-fs.com/training/ava_training_turns.htm advises that you pull back on the yoke: "The steeper the turn, the more elevator you’ll need. Watch your altitude!"

@DavidK You're absolutely correct. However, the subject matter being discussed here is not "what is the proper level-turn technique?" but rather the behavior of the angle of attack in a turn which may include a descent initiation even if it is fleeting.  This statement also assumes pilot technique causing negligible dihedral effect.

@RyanMortensen If you maintain exactly 1g total perceived gravity throughout a banked turn, I believe you're going to have more than a "fleeting" initiation of descent. I also think if you maintain the same elevator angle as in level flight, you'll end up in a downward spiral at increased airspeed, therefore increased lift and greater than 1g total perceived gravity.

@DavidK Yes, but if you only have 1g initially and then increase it (the AoA and G-force), it could in fact be a fleeting, initial 1g condition.  Once again, this question is not discussing, "what is proper pilot technique in a turn?" We are discussing how AoA varies with bank, and for the purpose of discussion, there was a contemplation of g-forces vs. AoA in a bank.  As I said, you are indeed correct that maintaining a 1g wing load with a bank would accelerate an aircraft downward, which is generally undesirable, but that is not the point of discussion.

@RyanMortensen The original question asked, "how does turning increase [AoA]?" I thought the context was a turn under control of a pilot. Shortly after Simon's last comment on this answer I think the discussion strayed, but I've already written too much, so I give up.

@DavidK I suppose what you're wondering is why we were bothering to discuss a 1g turn condition at all?  Obviously, increasing the load factor by performing a level turn will increase AoA.  The point of the 1g discussion was to concur that there is not anything else going on, by eliminating as many variables as possible when compared to a straight and level flight condition.  It was merely an analytical, not practical, discussion.

This good answer could be improved by pointing out that thrust and power must be increased to allow this to happen, i.e. to allow the airspeed to be maintained even after the angle-of-attack has been increased.

The question itself is open ended but, in the simplest terms, it depends on how you execute the turn - are you flying a constant airspeed or a constant angle of attack?
In either instance, in a level turn,  the load factor will increase with the inverse cosine of the bank angle.
If flying a constant airspeed turn, then an increase in lift coefficient will command a higher angle of attack as dynamic pressure remains constant: CL-turn = CL-wings level (L-turn/L-wings level)
If flying a constant angle of attack turn (i.e., same angle of attack in the bank as that used in level flight), the dynamic pressure has to increase to support the higher wing loading at a constant lift coefficient. Since dynamic pressure is a function of the velocity squared, the velocity commanded to maintain level flight within a banking turn varies with the square root of the load factor: V-turn = V-wings level x (load factor)^1/2   ...all other factors being equal.

How does the angle of attack vary in turns?

While many of the answers focus on the need for the angle-of-attack to be higher in a turn than in wings-level-flight if altitude is to be maintained with no change in power setting (or alternatively if altitude is to be maintained with increased power but no change in airspeed), your question seems to be asking about something else.

I read somewhere that when the aircraft banks to turn, its angle of
  attack increases.Since angle of attack is the angle between the
  relative wind and chord line, how does turning increase it? Is it
  because the relative wind is now from different direction or is it
  because the wing that goes higher meets the air at a higher angle?

You seem to have the impression that the angle-of-attack is inherently tends to increase as we enter a turn.
Of course when considering this whether this is true and why, an important question is "is the pilot in the loop, making pitch control inputs as needed to achieve some given goal, such as maintaining altitude without changing the thrust or power setting, or such as maintaining both airspeed and altitude (which will require more thrust and power)?
Many of the other answers to this question assume that the pilot is in the pitch control loop, manipulating the controls as needed to to achieve parameters such as the ones stated above.  In this case, the angle-of-attack usually will indeed be higher in the turn than in wings-level flight, for reasons given in other answers such as this one.
Of course, there are exceptions.  Here is an example of a situation where the angle-of-attack would ideally not be higher while turning than in wings-level flight, even when the pilot is in the pitch control loop: a glider is soaring in weak but smooth and widespread ridge lift, flying at the angle-of-attack that yields the minimum sink rate in steady-state linear flight.  In this case, as long as the roll rate is kept low, it would be a disadvantage rather than an advantage to increase the angle-of-attack when entering a turn.
But generally speaking, especially in the context of powered flight, a pilot will increase the angle-of-attack when entering a turn.  In fact, if you know you prefer to travel in some other direction than the one you are currently going in, in some cases you might want to do a little zoom climb as you roll into the turn, to convert some kinetic energy into altitude and get the turn accomplished quicker as well.
On the other hand, if we assume that the pilot is not in the pitch control loop, then your impression that the angle-of-attack tends to increase as we enter a turn is mistaken.  Here's why:
First of all, it's worth noting that if we start with the aircraft in a somewhat nose-up pitch attitude relative to the flight path-- i.e. a somewhat positive "angle-of-attack of the fuselage"-- than any rolling motion toward a steeper bank angle converts angle-of-attack into sideslip angle, so that the angle-of-attack tends to decrease.  This happens because an aircraft generally rolls about its longitudinal axis, not its velocity vector (flight path vector). So the angle-of-attack tends to decrease, as if we were pushing the control stick forward, while applying some top rudder at the same time. If this is hard to visualize, imagine starting at an extreme angle-of-attack, like 15 degrees or more, and rolling to very steep bank angle.  Since the aircraft's natural pitch stability dynamics are constantly trying to return the aircraft to the "correct" angle-of-attack for the elevator position (with one caveat that we'll note below), this dynamic will be most pronounced when the roll rate is very high and the aircraft's pitch stability dynamics are relaxed, i.e. the CG is so far aft as to give near-neutral pitch stability.  You aren't going to detect this dynamic as you practice maneuvers for the private pilot or commercial flight test in a typical general aviation trainer.
But what if we are not beginning the maneuver at a high "angle-of-attack of the fuselage", and/or we are rolling slowly enough that the above dynamic is trivial?  For example, what if we are making a typical turn entry in a general aviation light airplane, or perhaps a sailplane?  It turns out that if we leave the control stick or yoke in a fixed fore-and-aft position as we roll (or if we allow the control stick or yoke to "float" at trim in the fore-and-aft sense and apply no forward or aft pressure on it), we still find that the angle-of-attack tends to decrease as we increase the bank angle.  The reason for this is that as we bank, we start to turn, and the curving flight path actually creates a curvature in the free-stream airflow or relative wind, which tends to "push up" on the horizontal stabilizer and pitch down the nose, or more precisely, tends to alter the angle-of-attack of the horizontal stabilizer in a way that causes the aircraft to transition to a lower angle-of-attack overall, including a lower angle-of-attack as measured at the wing.  This effect is related to bank angle, not roll rate.
One way to see this effect first-hand in some aircraft is to trim for level flight with the stall horn just barely sounding, and then increase the bank angle without exerting any fore or aft pressure on the control yoke (or while using a special clamp to hold the yoke in a fixed position in the fore-and-aft sense while still allowing free roll control inputs), and note that the stall horn stops sounding when we slowly roll the aircraft into a moderately-banked turn, such as 45 degrees of bank.  In the turn, we are further from the stall angle-of-attack than we were in wings-level flight.  Be sure to try it both directions so as not to be misled by asymmetries in the angle-of-attack of the inside wing and outside wing during the turn.
Because of this effect, to maintain the same angle-of-attack in turning flight as we had in wings-level flight, we have to have the control stick or yoke further aft in the turn.  Since turn radius is proportional to velocity squared, this dynamic is especially pronounced in aircraft that fly at relatively low airspeeds and also fly at relatively low "scale speeds", i.e. take a relatively long time to cover one fuselage-length.  A sailplane circling in a thermal updraft would be a good example of such an aircraft, and it's not uncommon for a pilot in such a case to be circling with the stick so far aft that it would command a full stall if the wings were level rather than banked.
Another way to look at the curving relative wind induced by the turn, is to note that the aircraft is rotating around the pitch and yaw axes (and in some cases the roll axis as well), and this rotation induces a change in direction and magnitude of the local free-stream airflow as measured at various points along the length and span of the aircraft, just as different points on the blades of a pinwheel are not travelling through the airmass in exactly the same direction at any given point in time.  By this point of view, the tendency of the angle-of-attack to be lower when turning than when wings-level can be viewed as a natural consequence of a "pitch damping" effect, creating a resistance to rotation about the pitch axis which requires a change in elevator position to overcome.
Now to "throw one more wrench into the gears"-- the discussion of the effect of roll rate in the first paragraph of this answer referred to the combined average angle-of-attack of both wings.  It's simultaneously true that a rolling motion tends to increase the angle-of-attack of the descending wing and to decrease the angle-of-attack of the ascending wing.  This is important, as in extreme cases it can lead to a stall of the descending wing.  This is also an important cause of adverse yaw that it is not directly related to actual changes in the shape of the wing-- for more see https://www.av8n.com/how/htm/yaw.html#sec-adverse-yaw.  Note that even a constant-banked turn involves some degree of roll rate if the altitude is not constant with respect to the surrounding airmass-- the roll rate is toward the inside wingtip in a descending turn, and toward the outside wingtip in a climbing turn, and this will tend to create a difference in angle-of-attack between the two wings that tends to destabilize a climbing turn (i.e. makes the bank angle tend to increase) and stabilize a descending turn (i.e. makes the bank angle tend to decrease).  But that's probably more than you wanted to know!
Related ASE questions and answers--
This answer considers the consequences of the curving relative wind and the resulting nose-down pitch torque and decrease in angle-of-attack when the aircraft is maneuvering purely in the vertical plane-- e.g. during a loop.  Includes links at end to outside sources discussing the curvature of the relative wind in turning flight.

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What is a skidding turn (vs slipping turn)?
I am trying to understand the concept of

What qualifies a turn to be skidding or slipping

how we need to give elevator back pressure if we do not want to 'slip' the turn.

While browsing for answer I read that slipping occurs in the absence of back pressure. Why does it occur when there is no elevator back pressure. I understand that when you bank to turn, you reduce the lift component and not giving back pressure might make the aircraft nose low.
But how does that make a slipping turn?

Source: Wolfgang Langewiesche's 1944 Flight-dynamics-for-pilots classic text "Stick and Rudder"

Elevator backpressure is needed to maintain altitude and speed; I don't see how it should be related to slip or skid.

The question is a little confusing, are you asking the difference between a slipping/skidding turn, or are you wondering what role the elevator plays in making a turn skid or slip?

@RonBeyer: Actually both. I will update my question to make it clear. Sorry for that

where did you found that quote? that use of "slip to the side" has a different meaning from "slipping turn"

wolfgang langewiesche, stick and rudder

The terms "slip" and "skid" refer to two different types of uncoordinated turn - neither has much to do with the elevator, instead both depend on what the rudder is doing:

Skids
In a skid you have too much rudder input for the turn - the aircraft starts to pivot into the turn.

Because the aircraft is effectively pivoting about its center of lift while flying you can think of a skid as the outside wing advancing (seeing a faster relative wind), while the inside wing is retreating (seeing a slower relative wind). The inside wing thus generates less lift, and the plane starts to bank into the turn.
If you try to correct for that bank using aileron to raise the wing your elevator deflection effectively increases the angle of attack on that low wing, with its reduced relative wind -- do that aggressively enough and the inside (retreating) wing stalls while the outside (advancing) wing is still generating lift, and you enter a spin.

Slips
A slip is the same thing in the opposite direction: You have too little rudder input for the turn, and the aircraft is pivoting out of the turn:

In a slip the inside (low) wing is advancing, and the outside wing is retreating, so the inside wing is moving faster, generating more lift, and trying to level the plane. If you try to hold the inside wing down with aileron input you are increasing the angle of attack on the outside (high) wing to keep it up.  If the outside wing stalls because of this it will drop the aircraft into something closer to a wings-level attitude, which will also bring you out of the turn.
This is why slips are said to be spin-resistant: The stall encourages the aircraft back into a more stable configuration (coordinated flight) rather than a less-stable one (spin).

The images here are taken from BoldMethod's "Why skids are more dangerous than slips" article, which is a great read and also has a better and more comprehensive explanation of the aerodynamics involved.
My explanation above is a vast oversimplification, but the advancing/retreating wings and difference in relative wind is the one that got the idea to stick in my head so maybe it will help you out too.

Excellent, vortaq! I've often wondered what the difference was, but never bothered asking. Now I don't have to.

Can we use these Boldmethod images here? Those images are protected by copyright, we don't seem to have a license for them, and this doesn't look like "fair use" to me.

It seems that this answer is intimately connecting rudder deflection to yaw rate, and therefore to roll rate.  I'd submit that rudder deflection is more closely connected to sideslip /skid  angle, and therefore to sideforce and slip/skid ball deflection. Sure, in some circumstances, such as when you are entering a spin, rudder deflection strongly affects yaw rate, but that's not the general case.

You can also fly in a straight line while in a side-slip. eg right rudder with enough left stick and bank to stop any turn. Drag is high because the aeroplane presents one whole side (fuselage and fin) to the relative airflow.
This method is used to control approach angle when the aeroplane has no flaps (or ineffective ones) and is as old as aviation itself. It also allows the runway to remain in the pilot’s view in aeroplanes with limited forward visibility like the Pitts Special biplanes.

(1) When you turn with no rudder there is natural tendency to skid slip, slide away into the center of the turn. The tail swings in and nose goes out. This is bad because the inner side of the plane (the side inside the turn) hits the air and you get drag.
(2) To counteract this tendency, the pilot applies rudder to pull the tail nose in. The whole idea is to keep the fuselage in line with the direction of motion and prevent undue drag.
(3) In any turn, without making a correction, you will lose altitude. This is because when you bank, less profile is facing downwards so your resistance to gravity decreases. Imagine if you banked to 90-degrees, what would happen? You would plummet like a rock. The same thing happens whenever you bank, you decrease your ground-facing air resistance. There are two ways to counteract this: pull the stick back and/or apply power. The general practice is to put back pressure on the stick. This helps you maintain altitude at the cost of airspeed. A better practice is to increase power. By increasing power slightly you can maintain altitude in a turn without sacrificing speed. In any low-altitude maneuver, for example, when taking off, you should always increase power when turning, if you are not at maximum power already.
(4) Back pressure on the stick has nothing to do with skidding or slipping. It is purely a move to maintain altitude during a turn.

when you say slide 'away', does it mean flight goes inside the turn radius or outside?

@user2927392 "away" always means away from the center of the turn. A skid moves away from the center of the turn, a slip moves towards it.

I think your answer conflicts with the accepted one. You are saying when you turn with no rudder you skid, tail swings out, but the accepted one says your tail would stay closer to the center.

@Hot.PxL I don't see where the accepted answer makes any comment one way or the other on a turn with no rudder input, however I can tell you this: if you turn in a small aircraft without making any rudder inputs, then the plane will tend to skid.

@TylerDurden My understanding is, for example we turn the yoke and try to enter a turn, but apply no rudder. Then the nose will want to go away and the tail will stay inside the (imaginary) circle which CG glides through. (The "slip" image in accepted answer is depicting). Is that correct?

@Hot.PxL Yes, you are right. What I said before is wrong. The nose goes out and the tail goes in, so it is indeed a slip. Sorry about that, got confused.

A skidding turn is an uncoordinated turn causing the plane to skid to the outside of the turn. A turn in an airplane is accomplished by shifting some of the lift from up to a vector in the desired direction. The rudder is used to help coordinate the turn. During the turn there is more lift on the outside wing than the inside wing. That creates a yawing effect that the rudder is used to overcome. There are times when a very small direction change can be accomplished using just the rudder. But primarily a turn happens because lift is vectored to one direction or the other. That is why when you turn a certain amount of back pressure is required on the elevator. You are removing some of the lift from up, to right or left. The steeper the turn, the more upward lift you are sacrificing for more right or left direction.
A slip is the opposite of a skid. It is an uncoordinated turn wherein the airplane slides into the turn.
In either case, it is the rudder that is used to overcome the adverse yawing effect of a turn, not the elevator. The elevator is used to overcome the loss of upward lift.
Uncoordinated slow speed turns during the landing sequence are probably the most common reason for small plane crashes. The point when the airplane turns from the base leg to the final leg of landing is, in my opinion, the most dangerous part of flying. An uncoordinated slow steep turn is an invitation to a spin which happens too low to the ground to overcome.

There's really not much point in trying to answer this question, without first pointing out that the content you have quoted at length from Langewiesche's classic text "Stick and Rudder" appears to be erroneous.  In actual practice, pitch inputs generally have only a very small effect on yaw "coordination" (keeping the ball and yaw string centered).  For more, see this answer to the related ASE question "Why does elevator input move the turn coordinator ball in steep turns?"
But in short--
Skidding turn-- ball toward high wingtip (and toward outside of turn), yaw string toward low wingtip (and toward inside of turn).  Add "outside"/top rudder to coordinate.
Slipping turn-- ball toward low wingtip (and toward inside of turn), yaw string toward high wingtip (and toward outside of turn).  This is what normally happens if we roll the airplane into a turn without using the rudder.  Add "inside"/bottom rudder to coordinate.
This situation is also considered to be a "slip" or "sideslip" -- linear flight path,  ball toward low wingtip, yaw string toward high wingtip. If this an intentional maneuver, pilot is undoubtedly holding "top" rudder, and "bottom" aileron.1 Adding "bottom" rudder is one way to end the slip.
This situation is arguably ambiguous as to whether it should be considered a "slipping" turn or a "skidding" turn-- rudder hard over toward high wingtip, flight path curving toward high wingtip, ball toward low wingtip (and toward outside of turn), yaw string toward high wingtip (and toward inside of turn).
Footnotes:

The degree of "bottom" aileron required to neutralize the net roll torque in this situation, and in any other slipping or skidding situation, is highly dependent upon the slip-roll coupling characteristics of the aircraft.  In the vast majority of airplanes, some "bottom" aileron would be needed, but in a mid-wing aerobatic airplane with zero dihedral, little or no "bottom" aileron might be required.

A full answer really ought to talk about the role of aerodynamic sideforce from the fuselage flying sideways through the air-- how it affects the slip-skid ball, and the turn direction and rate. Plenty of related ASE questions/ answers could be linked to address this.

But-- this is enough for now.

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/


text:
Viable to slip onto the center line?
One of the constant problems of aviation is drawing onto the center line on final. Either due to a bad rollout or due to crosswinds a pilot often finds that the aircraft is not on the center line for landing. If the aircraft is far off the line, we can end up doing buttonhook  turns and other dangerous maneuvers to get on line.
I am wondering about the viability of side slipping onto the line instead of doing banking maneuvers. In other words, there would be no bank and no turn. You would side slip and add power to maintain altitude, thus theoretically that would just slide the aircraft onto the line. This seems like a potentially more reliable and controllable way to adjust an aircraft onto the center line during final. What is the viability of this maneuver?
It seems to me there is a logic to it, because when I am dealing with heavy gust conditions this is basically what we are taught to do anyway: you put the windward wing down and side slip into the crosswind as you land to keep a level attitude. So, we are often doing this maneuver anyway, why not use it adjust to center on final?

Yes, but somehow it's OK to do it when there is a crosswind component?  Really depends what you're flying.  I was actually waiting for a 3 degree powered approach with huge wings to chime in.  There one can yaw and use the thrust component to hold line.  This helps avoid wing strike.  Flying a good pattern is important in all cases.

Sideslip will not move the aircraft laterally. It only changes the direction that the nose is pointed in, it does not initiate a turn (a change in the direction of the flight path/ground track). So by itself, it cannot put your ground track on the center line if it is not already there. All it can do is align the nose direction the nose is pointed in (and. more significantly, the main gear wheels and tires) with the centerline. That is only one part of the problem. You also need to ensure that the ground track is aligned with, and over, the centerline.

Flying the aircraft so that it is over the runway centerline is simple. When the runway, (and the centerline) appear to be absolutely vertical in the windscreen, you are over the extended centerline. Aligning the aircraft ground track with the centerline is accomplished by keeping it that way. If the runway centerline "picture" twists so that the near end is below and moves to the right, you are drifting left, you need to move the aircraft back to the right. If the near end is below and moves to the left, you are drifting to the right, and you need to turn the aircraft and move right.

@CharlesBretana  please review "side slip"  and "forward slip".  "Side slip" is the one for lateral movement.  Remember, lift is by far the most powerful force an aircraft produces, why not use the horizontal component to move sideways?  (banking).

@RobertDiGiovanni The most powerful force an aircraft produces is the force ejecting money from your wallet.

"you put the windward wing down and side slip into the crosswind as you land to keep a level attitude."  -- what is the meaning of "to keep a level attitude" in this sentence?  Would the phrase best be removed?  Are you talking about bank, or pitch attitude, or ??

Really I suspect that most pilots are not fully aware of exactly how they control an aircraft in the last few tens of feet of altitude before touchdown

Related: https://aviation.stackexchange.com/questions/88442/how-much-slipping-and-sliding-go-on-during-short-final

I’m not totally sure how you’re orienting that, are you talking about aligning the aircraft up on final approach from a base leg, then doing a slip to final alignment on the runway CL for final approach?
The best way to get the aircraft aligned on runway centerline during the base to final turn, is to note where the runway will centerline will intersect the base leg flight path if it were extended out that far, thence plan your turn from base to final to roll out on that imaginary centerline without being rushed or overshooting.  In addition a pilot may opt to fly a crabbed final approach as opposed to a side slip.  As the pilot enters the round out, they should use rudder pressure to align the long axis of the airplane parallel to the centerline and aileron pressure to hold the airplane on centerline with the main wheels straddling the CL.
NOTE WELL:  if your airplane over shoots or under shoots runway centerline during the base to final turn, you feel rushed in the turn, or have to use some sort of gross maneuver on final to get the aircraft position back on center line, you should immediately perform a go around as opposed to continue to force the airplane to fly final in land.

"As the pilot enters the round out, they should use rudder pressure to align the long axis of the airplane parallel to the centerline and aileron pressure to hold the airplane on centerline with the main wheels straddling the CL." -- just curious-- are you talking about only in the presence of a definite crosswind, or also just to counteract small random sideways drifts or pilot errors/ oversights-- say you are landing in calm winds and as you are just starting to round out from final approach you notice you are 3 feet left of centerline, do you use this technique?

If so it seems the answer to original question (apart from the part about zero bank angle which is not possible) would be "yes, at least for small corrections near the ground".

Side slipping to the center of the runway is an excellent technique for high wing aircraft (with minimal risk of a wingtip strike).  One might correct a base to final overshoot with a coordinated turn.  Once on the final glidepath, finer adjustments can be made by sideslipping.
Generally, it is not a good idea to combine uncoordinated flight with a turn, but the side slip is much safer than skidding with excess rudder.
Adequate airspeed, and practice at altitude, is essential in keeping low and slow manuvering safe.  One can stay coordinated, and not slip until after rounding, and some prefer to side slip (using opposite rudder) higher up on final once the aircraft has completed its turn. This has the advantage of lining the nose up with the runway earlier. Others come in on a circle, controlling rate of descent by side slipping (varying bank angle).  So, one can side slip with more power for lateral movement, or simply "drop a wing" to bleed off a few feet of altitude.
The technique is known and widely used.  Safety will depend on knowing the limitations of the model of aircraft being flown.

Viable to slip onto the center line?

Yes, we can use a sideslip to shift the aircraft sideways relative to the runway, without changing heading.  More on this below.  However, the body of the question contains some misconceptions--

I am wondering about the viability of side slipping onto the line
instead of doing banking maneuvers. In other words, there would be no
bank and no turn.

This excerpt shows a misunderstanding of basic flight dynamics. The fundamental characteristic of a sideslip is that the aircraft is moving sideways through the air.  As an aircraft enters a sideslip, there's generally going to be a bank, a turn, or both.
If rudder is applied and held while the ailerons are manipulated as needed to keep the wings level, the aircraft will fly sideways through the air (sideslip), and the aerodynamic force resulting from the air striking the side of the fuselage will cause a curvature in the flight path-- a turn-- in the direction of the deflected rudder.  So now we're actually in a skidding turn.1  The aircraft's heading will change.  You could steer the aircraft closer to the runway centerline this way, but why would you?  Much better to turn the normal way-- by banking.  Skidding turns are generally considered to be bad airmanship, and an invitation to a spin.
On the other hand, when we talk of "sideslip" we often envision a maneuver where the aircraft is kept in a bank opposite the direction of the rudder input.  For a typical "forward slip" to increase the descent angle and sink rate, we would apply rudder, moving the nose to a different heading and exposing the side of the fuselage to the airflow, while simultaneously applying bank as needed to provide a sideways force that cancels the aerodynamic sideforce from the sideways airflow striking the fuselage, preventing the flight path from curving in the direction of the applied rudder.  So we're now flying sideways through the air, but the direction of the flight path and ground track hasn't changed-- this won't bring us any closer toward the runway centerline.
Alternatively, we could first apply the bank angle, while manipulating the rudder as needed to prevent any heading change at all.  This is the typical "sideslip" that we do to compensate for a crosswind on final approach.  Aerodynamically, for any given bank angle, this kind of "sideslip" ends up being exactly identical to the "forward slip" described immediately above-- we've just entered it slightly differently, giving the flight path and ground track some time to change direction (toward the low wing) before the full rudder correction is applied.2  So we've actually accomplished a turn as we entered the maneuver-- we've angled or "bent" the direction of the flight path and ground track toward the low wingtip.3  Once the maneuver is stabilized, with constant bank angle and rudder deflection and heading, the flight path and ground track are linear again, but no longer aligned with the aircraft centerline (at least in the no-crosswind case).  The fuselage ends up moving sideways through the air, like in any other sideslip.
We could use this maneuver to move toward the runway centerline without changing heading.  (Not to move toward the runway centerline without banking.)  And for small corrections, this may be in fact what pilots sometimes (perhaps unconsciously) do, especially when very near the runway.  Of course, the wings should be leveled again (in the no-crosswind case) to stop all sideways "drift" before touchdown-- touching the ground while moving in a different direction than the aircraft is actually pointing can be hard on the landing gear.  For larger corrections, carried out higher from the ground, it's generally better just to do a shallow turn toward the centerline, "coordinated" in the usual manner.
For completeness, we should also note that in the case described immediately above, when we remove the bank and the rudder input to exit the sideslip, without allowing the aircraft to change heading, we again accomplish a "turn"-- we "bend" the direction of the flight path (and the ground track, in the no-wind case) so that they are again parallel to the direction the aircraft's nose is pointing.  Technically, this could be described as a very slight "skidding" turn-- as we are taking out the bank angle, we delay taking out the rudder input for just long enough to allow the sideforce from the air striking the side of the fuselage to "bend" the flight path and ground track in the direction of the deflected rudder (i.e., toward the high wingtip.)  You'll never see this maneuver analyzed in this level of detail in any flight training materials--and knowing this extra detail won't make your landings any better-- but that's in fact what is going on.  From the pilot's viewpoint, however, it's simpler just to focus on the fact that by banking the wing, without allowing the heading to change, the aircraft is being "slid" sideways toward the low wingtip, and this "sliding" ends once the bank is removed.4
Footnotes--

The language gets kind of complicated here.  To an engineer, any sideways motion of the aircraft through the air is a form of a sideslip.  To a pilot, if the sideways motion is toward the high wing, or if the wings are completely level, the maneuver is called a "skid" rather than a "slip" or "sideslip".

This, in a single sentence, is the slight difference between the control inputs used to enter the maneuvers commonly described as a "sideslip" and a "forward slip".  Pertains to many other ASE questions -- such as this one: What is the difference between a forward slip and a side slip? .  No suggestion is intended here that a pilot is typically aware that he or she is doing something slightly different with the rudder when entering a forward slip versus a sideslip-- it's just a natural result of the pilot's intention to either keep the aircraft's heading constant during the entry, or to cause it to change.

Of course, the "turn" we're talking about here is quite modest-- for the aircraft heading to stay constant, the direction of the flight path through the airmass must "bend" or change by an angle that is exactly equal to the angle between the direction of the flight path through the airmass and the direction the centerline of the aircraft is pointing.  I.e., by an angle equal to the yaw string deflection, in the case of a "perfect" yaw string mounted on a long probe well in front of the nose of the aircraft.

This simplification-- focussing on the sideways "sliding" aspect of the maneuver while ignoring complete dynamics of the actual turns that take place as the maneuver is initiated or terminated-- arguably has tinges of an Aristotelian rather than Newtonian understanding of physics-- one can easily draw the incorrect conclusion that a steady net force is required to cause a steady linear motion.

Really I suspect that most pilots are not fully aware of exactly how they control an aircraft in the last few tens of feet of altitude before touchdown

Every one please read footnote 1.

Sideslip will not move the aircraft laterally. It only changes the direction that the nose is pointed in, it does not initiate a turn (a change in the direction of the flight path/ground track). So by itself, it cannot put your ground track on the center line if it is not already there. All it can do is align the direction the nose is pointed in (and, more significantly, the direction that the main gear wheels and tires are pointed in) with the centerline. That is why we use this for crosswind landing, to minimize the sideload forces on the landing gear.
But that is only one part of the landing problem. You also need to ensure that the ground track is aligned with, and over, the centerline. Flying the aircraft so that it is over the runway centerline is simple. When the runway, (and the centerline) appear to be absolutely vertical in the windscreen, you are over the extended centerline. Aligning the aircraft ground track with the extended runway centerline is accomplished by keeping it that way (vertical). If the runway centerline "picture", as you fly down final, twists in the windscreen so that the near end is  moves to the right, you are drifting left of centerline, you need to move the aircraft back to the right. If the near end moves to the left, you are drifting to the right, and you need to turn the aircraft and move right.
In practice, if you consciously look for this, and monitor it, this is absurdly simple and easy. If you are not looking for it, you will not be aware of it.

Comments are not for extended discussion; this conversation has been moved to chat.


text:
Do you need an instrument rating to fly in IMC in Class G airspace?
I’ve been watching IFR checkride prep videos and several of them mention that among the things you can do with your IFR ticket is to:
Fly in Instrument Meteorological Conditions (IMC) in controlled airspace.
If you have your IFR ticket, you can depart Class G in IMC if you have filed a flight plan. So I’m a bit confused about why they put that restriction on the privilege.
But it also got me thinking. You need an IFR ticket to fly in IMC in controlled airspace, but is there a regulation that says you need one to fly in IMC in uncontrolled airspace? Or does NTSB 3935 mean that flying in IMC without your IFR ticket automatically imply careless and reckless flying?

You cannot fly in IMC in any airspace without an IFR rating. What qualifies as IFR changes with the airspace according to the VFR rules. G just has the most relaxed conditions. Usually flying at minimums at low altitude is called "scud running" and greatly reduces your lifespan.

@RonBeyer I can’t find any regulation that says that you can’t fly in uncontrolled airspace without an IFR rating. Can you point me to the regulation?

RonBeyer means when IMC.  It is illegal anywhere, regardless of the class of airspace, to fly in IMC without a rating and appropriate aircraft since the requirements for VFR  cannot be met.

It's pretty simple, if you aren't VFR, you are IMC, which requires an IFR rating.

@JScarry, reference 14 CFR 61.3 (e)  (e)Instrument rating. No person may act as pilot in command of a civil aircraft under IFR or in weather conditions less than the minimums prescribed for VFR flight unless that person holds:

@mongo That’s what I was looking for. I thought it should be in Part 91 Instrument Flight Rules and it didn’t occur to me that it would be Part 61.

@JScarry As a general rule, 91 deals with operations and equipment, and 61 deals with people and qualifications.

My edit changed the meaning pretty substantially, but I think that's what you actually meant to say.  Please roll it back if not.

Re "If you have your IFR ticket, you can depart Class G in IMC if you have filed a flight plan." -- apparently some ASE contributors would dispute this -- see for example https://aviation.stackexchange.com/questions/21888/can-an-ifr-clearance-be-issued-and-flown-through-imc-in-class-g-airspace/74345#74345

Also re "If you have your IFR ticket, you can depart Class G in IMC if you have filed a flight plan."-- and some ASE contributors would dispute that it is even necessary to file a flight plan at all in such a case as long as you will not violate VFR cloud clearance and visibility minimums by the time you reach controlled airspace-- see for example https://aviation.stackexchange.com/questions/21888/can-an-ifr-clearance-be-issued-and-flown-through-imc-in-class-g-airspace/31317#31317

In the US, an appropriate instrument rating is needed to fly when WX conditions are below VFR minimums, and at night for SVFR.  14 CFR 61.3(e)
Additionally, one needs an appropriate instrument rating to fly under an IFR clearance when in controlled airspace, regardless of the WX.
Notes: ATP and things like a category and class in an airship are exceptions. Also an airplane instrument rating works in gliders. Etc.
Addendum: Specific to the OP's citing a NTSB report, the following comment is added: To rely on the mentioned NTSB report, it is important to read the findings, and understand that, a.) the pilot intended to hit VFR conditions before 700AGL and upon entering controlled airspace, b.) the pilot told people he was taking off despite qualified observers noting 200 ft clouds, c.) the airport where approach was based said they were below IFR, which would leave the pilot few options if he needed to land nearby. Given that, it is easy to agree with the NTSB. Just because something is legal, doesn't mean it is best practice, and in this case it is likely the pilot did not do as stated.
One final point, IFR without a clearance in Class G is a decision which should not be taken lightly. Years ago, I routinely did it, and the airspace was uncontrolled at safe IFR enroute altitudes.  Departing a Class G airport, with less than VFR conditions, and expecting VFR before hitting controlled airspace at 700AGL is dubious. The vest majority of my IFR flights in uncontrolled airspace, that were without benefit of clearances, were where there are large areas of uncontrolled airspace, and safe IFR altitudes could be flown enroute. Furthermore, while ATC did not issue clearances for flight in uncontrolled airspace, coordination with ATC where possible, and also with local FSS were done. That is different from popping off an airport that is likely IMC, and stating that one intends to climb into VFR weather before hitting 700AGL and controlled airspace. The frosting on the cake is having the field in the area be below IFR limits for approaches.  Good judgment is hard to legislate.

Generally, the FAA considers exercising privileges which one is not certificated in as evidence of careless and reckless operation.

If you are flying in IMC in uncontrolled airspace you are not IFR, you are just IMC in violation of class G cloud clearance requirements.  Skud running is still VFR.

@MichaelHall, Not sure I understand your comment.  If IFR rated and equipped, one may fly IMC in class G airspace, consistent with the rules.  Because the airspace is uncontrolled, there are no "clearances" but generally ATC will provide coordination.  Today there is much less uncontrolled airspace, and this is less commonplace in the US.

The videos you were watching did not mean to imply that you can fly in IMC in uncontrolled airspace without an IFR rating.  Rather, it seems that the creators of the videos wanted to avoid suggesting that you can fly in uncontrolled airspace in IMC with an IFR rating.
Some pilots believe that no legal flight in IMC is possible in uncontrolled airspace.  (See for example this ASE answer).  This is simply not so.  If it were, why would some of the approaches into KCVO, which has a Class E floor at 700' AGL, have "minimums" well below that that?  (Click here to download approach plates.)
In fact, it is legal for a pilot with an IFR rating to fly in IMC in uncontrolled  airspace, at least under some circumstances.  However, there is no provision for adequate separation from VFR traffic, which is allowed to fly right up to the edges of the clouds in uncontrolled airspace, at least below 1200' AGL in the daytime.  This strange discordance in the FARs had led the FAA to use FAR 91.13, the "catch-all" regulation against "careless or reckless operation", to sanction some pilots for operating under IFR in uncontrolled airspace in some circumstances, as described in this 1993 NTSB ruling, also referenced in this 2016 Letter of Interpretation from the FAA.
Evidently the creators of the videos you were watching wanted to spare you the possibility of sanction, persecution, or prosecution by the FAA or NTSB, and so they chose language that avoided any suggestion that an IFR rating was a blanket license to operate in IMC under IFR in Class G airspace, and chose not to "muddy the waters" by letting you know there are in fact many situations where the FAA does condone operation in IMC under IFR in Class G airspace.
If the FAA had ever at any point made it clear under exactly what circumstances a pilot may operate in IMC under IFR in uncontrolled airspace without fear of sanction, the creators of the checkride prep videos would undoubtedly have included that information in the videos, but unfortunately this is not the case.
This answer begs another question which has not yet been asked on ASE, though it is addressed in some of the links below: "May an IFR-rated pilot flying an IFR-rated aircraft in Class G airspace violate the VFR cloud clearance requirements without an IFR clearance?"
Related ASE answers --
Can an IFR clearance be issued and flown through IMC in class G airspace?
Can an IFR clearance be issued and flown through IMC in class G airspace?
Can an IFR clearance be issued and flown through IMC in class G airspace?
Do you need an instrument rating to fly in IMC in Class G airspace?

Lovin' the DV's as always...


text:
Was the 727 the last airliner whose doors could be opened in flight?
The Boeing 727, put into service in 1963, had the distinction of a rear doorway with stairs that could be lowered so that passengers could embark and disembark at rural airports which did not have their own airstairs.  As initially designed, this door could be opened even whilst in flight as it is not the usual "plug" style.
In 1971, the hijacker known as DB Cooper famously used the rear door to make his parachute escape mid-flight. According to wikipedia, the CIA also used this feature to drop passengers and supplies mid-flight.

In the wake of the DB Cooper incident and other hijackings, the FAA mandated in 1972 that cooper vanes to be installed to prevent the opening of the rear door while in flight.
Does this make the pre-cooper 727 the last passenger airliner which had a door (either passenger or cargo) that could be normally opened while in flight without first depressurizing the entire cabin?
Small Print: There was a rear cabin, pressurized bulkhead door on the 727 that you had to transit through to get access to the ventral stairways which was in the aft, unpressurized bulkhead. The actual ventral stair door itself was not a "plug" style. Thus the ventral stairways meets the requirement of being able to open in flight (even if you couldn't get there because the rear cabin bulkhead door might not open).
Animated image from wikipedia and cc.

One would think they'd be glad to have a way to get a hijacker off the plane mid-flight...

I think you are missing the point that in the Cooper incident, the aircraft was at a rather low altitude when the rear door was opened (approx 10,000' as I recall), and Cooper may have instructed the pilots not to pressurize the a/c at all.

I love that animation. It's just... perfect!

In the Cooper incident, the aircraft was at a rather low altitude when the rear door was opened (approx 10,000' as I recall), and Cooper undoubtedly instructed the pilots not to pressurize the a/c at all.   The only reason that it's significant that the door was not a "plug" type design, is because a "plug" design could never open outwards and thus would be incompatible with an "airstairs" feature.  A "plug" type door could also have been opened in (unpressurized) flight, but would be less attractive than an "airstairs" door to someone planning to jump from the a/c.

The door of an airplane can be opened in any of the aircraft only when the airplane is depressurized. When an airplane is pressurized, the extra pressure inside the hull of the plane forces the door outwards which in turn presses on a seal and hence, to open a door in such a configuration, where 6 p.s.i. (pounds per square inch) of pressure is acting against several square feet of the door, one would need a tremendous amount of force, and even if the door was to be opened electronically or hydraulically, one would still expect rapid decompression of the cabin after opening the door.
Hence it is categorically incorrect to claim that modern (new) airplanes can't open their doors mid flight. It is quite achievable, but after depressurizing the cabin. Here is a source for your further reading.
Edit
The following airplanes have Air-Stair installed. I'm not sure which of them has an over-ride to open this Air-Stair mid-air, but any airplane which does not have a cooper vane installed (Or any other external barrier) can open the Air-Stair midair (technically):
Lockheed L1011
Boeing VC 25
Ilyushin Il-86
DC-9/MD-80/MD-90/B-717
Boeing 737
Some modified Airbus and Boeing planes have Air-Stairs too.

You're exploiting a loophole in my phrasing of the question (which I  should close). I don't think the 727 required depressurization to release the rear door. Thus it remains the only plane that doesn't need to be depressurized to allow a door to open.

It did not need depressurizing because the exit bay door opened outwards instead of inwards (And that's why cooper vanes were installed). Also, I did not exploit the loophole, but I answered to what your question asked (Can the doors be opened? YES they can be)

I amended the question to close that loophole.

Amended the answer

If memory serves, the rear stair control handle on the 727 is behind the aft cabin door in the unpressurized area. You have to depressurize the cabin to open the stairs.

You don't have to depressurize the cabin to open the stairs. You have to depressurize the cabin to open the rear cabin door that leads to the stairs.   If you're hiding in the area between the rear cabin door and the stairs, you can open the stairs without depressurizing the cabin.

@RoboKaren -- re " If you're hiding in the area between the rear cabin door and the stairs, you can open the stairs without depressurizing the cabin." -- but Cooper wasn't.

@VictorJuliet -- Re "It did not need depressurizing because the exit bay door opened outwards instead of inwards (And that's why cooper vanes were installed)." -- it's true that an "airstairs" door, which opens outwards, is incompatible with a "plug" design, which must open inwards, and cannot be opened when the aircraft is pressurized.  But you could also say that it did not need depressurizing because the jump was made from a relatively low altitude, and Cooper undoubtedly instructed the pilots not to pressurize the aircraft.

In fact an "airstairs" door would probably leak enough to prevent pressurization any time that there was not a "plug" type door installed between aft end of the cabin and the "airstairs" door.

There are more that have stairs installed, including the Fokker F-28, F-70, and F-100 (all rather rare birds these days, sadly).


text:
What is the minimum number of people required to manually pushback a 747?
Similar to :
Is there a minimum number of people needed for an aircraft pushback?
After watching this video I wonder what would be the number of people required to manually push back a 747.

I know small aircraft is usually pushback by the pilot and if possible a wing walker. Large jet at large airport has a few people involved in the process. But manual pushback a large jet is a special event where you need lots of people. How many do you need?

There are too many factors when considering humans to count, this might make your question slightly unanswerable.

I fail to understand why this is not a dupe. You may highlight why and how your question is different.

@ManuH one question is talking about manual pushback with no machines, the other question is asking about minimum number of people spotting etc to safely push back with a tug / tractor / whatever.  Not obvious from titles but more clear if you read the actual body of the questions.

@quietflyer such comment should be included in the question

@ManuH -- "But manual pushback a large jet is a special event where you need lots of people. "

The rope must be attached one end to the aircraft and another to some stationary point. Humans should push it onto the side from the middle point. This should amplify the force significantly, reducing the number of people required by at least the factor of ten or about, so I think somewhat all passengers could do, even if half of them is not fit (there are about 400 in three class layout and we need about 200 strong enough humans). Here is the description of the idea with all mathematics involved.

@vasin1987 and h22--  It seems fair enough to assume a rope and a stationary anchor point are available, I guess.  But why not assume some pulleys are available as well?  Just a thought...  Really I guess I am pointing out an ambiguity in the question as much as anything else.  In a typical pushback situation no stationary anchor point in a suitable location may be available.

Drawbar pull for 747-8 is 31,100 kg.
Humans can safely push 20% of their body weight.
So you need 155,500 kg of humans.
If we assume 80 kg per human and round up we have 1950 humans.
This number is for using only your arms, so using whole body, as would probably be the case here, you can double it.
So 975 (1950) humans to push with whole body.
Pull is more efficient at 30%, so only 103667 kg of humans needed.
This adds to 650 (1300) humans of 80 kg rounded up with whole body.
No idea if this is actually true, although numbers are (barely) low enough that someone might have tried. The big practical issue is connecting all those humans to the plane.
Also, the actual numbers to move the plane at all should be significantly lower. This is the number of people needed for safely operating the plane in normal airport conditions including slight upward slope and some engine thrust resistance.
If you only need zero slope and engines off, you should be able to cut the numbers by more than a third. That you are happy with much lower speeds and distances than Boeing should cut the numbers further.
Also, what I knew and intended to mention but forgot. These are the safe numbers for constant and repeated work intended to avoid workplace injuries. Peak exertion for a one time stunt is something else. I am providing these numbers as a starting point because the exact circumstances are not mentioned.

Where does the figure of pulling only 30% of your weight come from? I'm not an athlete, but I'm able to lift my own weight rather easily. If you look at truck pulling one person is pulling 30-40 tons... I would guess your answer is off by an magnitude.

@vidarlo, the number is probably true in the sense that the force applied is 20% of the person's weight (weight is a force and is measured in N, not kg). However to pull a 40 t truck (weighing 390 kN on Earth), straight and level, you only need something on order of 5 kN. After all the towbar used to push back aircraft weighing up to 4400 kN can handle mere 305 kN.

Hm, checking the linked diagram (page 78) it really seems like its saying you need ~170 kN to get the plane moving when fully loaded (the 305 kN figure is when the engines are already running at idle, which they shouldn't with people around). It still sounds like too much. In the truck pulling video the guy is not inclined even by 60°, so he is not pulling even twice his weight (he's helping himself with the rope, but that's relatively lose,so 2.5× max), which would be less than 1% of the ~355 kN weight of the truck…

… Roller bearing are roller bearings and should have similar coefficient of friction and the higher tire pressure of the aircraft should result in less resistance, not more. So the about 4% (~170 kN, the engines stopped case) still seems to be rather much.

@vidarlo I looked it up and the number is actually correct. It is just the wrong number. It was for using just your arms. Using whole body as would be the case here should roughly double it. So off by factor of two. The main confusion stems from the fact that it is not the number people commenting expect. This is the safe number not the maximum exertion. I'll edit to clarify that. I realized the confusion this might cause when I wrote the answer but forgot to mention it and just mentioned the issue with Boeing numbers explicitly. Oops.

@JanHudec The number Boeing gives is what the towbar needs to be able to take. It is probably based on the static friction with some unstated safety factors for stupid people, bad maintenance, bad manufacturing quality control, bad wind direction etc. People towing trucks use their mass and inertia to defeat the static friction.

I challenge the 20% of body weight figure.  I regularly push & pull my Piper Cherokee (1230 lbs empty, plus up to 300 lbs or so fuel) without undue effort, and I assure you that I don't weigh anywhere close to 3 tons :-)

@jamesqf It is NOT the weight of the object, it is the force you are pulling with that matters. A 747-8 weighs quite a lot more than 30+ tons. This was obfuscated since Boeing gave the number in kg and I was comparing it to portion of body mass in kg.

@Ville Niemi: My comment has nothing to do with whether the mass is in kg or not, since you're stating the limit as a percent of body weight.  It is that you're saying there's a 20% limit on the weight a person can push, when I expect you meant the force they can exert.  (That is, to move my Cherokee around the ramp, I probably exert only a few tens of lbs/kg of force, not the entire weight of the plane.)  I'd also quarrel with the 20%, since I can easily bench-press (or otherwise lift) well over my body weight, and do many reps at that weight.

@jamesqf Yes, it is the force they can exert and it is assumed to be proportional to the body mass, so that if necessary numbers are both given in the same units, here in kg because I speak metric, you can just take 40% or 60%. For your second point read the last paragraph I edited in yesterday after I realized I just plain forgot to write it.

@Ville Niemi: Well, I still have to disagree on the numbers.  In any case, based on my experience of moving light planes around, and doing things like push-starting cars (which with the several British sports cars I've owned, was perhaps more common than having a working starter), the force you can apply is most often limited by when your feet start to slip.

@jamesqf Definitely so. Since the number I am including is deliberately low (and no, I do not claim it is somehow more correct than other choices, just the one I made) it is not an issue for my answer but if you look at the pdf I linked you'll see that needed traction wheel load on the surface is also defined in table.

Looking at that video in full-screen 4K (it's still pretty blurry), I generously counted 13 people who looked like they had their hands on tires and/or wheel bogeies. So I'd say that it takes 13 people to push a 747 with engines running.
There are many additional people standing around, so there may have been more involved in getting it moving (the plane is already moving when the video started), so at a minimum, it takes 13 people to keep it moving.
There is no indication of how heavily loaded this particular plane is with pax, luggage/cargo or fuel, so this number may vary.

In main aeroports(no aerodrome) the pushback is done by the push(There are 2 pushback procedures: with clamp or towbar.). I don't think it has ever been done by men so we can't answer your question (except theoretically using formulas)
maybe :W = F × D × cos (θ), where W corresponds to the work in joules (J), F to the force expressed in newtons (N), D to the distance in meters (m) and θ to the angle between the force and direction of the object's trajectory.(for know the force). This formula is just to know the
the work of a force.i don't know the max of the work of a human's force so i can't divide and have the result.
If the intensity of the force is not known, it can be calculated from the mass of the object and its acceleration, assuming that there are no other forces acting on it. . The formula to apply is as follows: F = M × A.
we don't take this into account:In physics, work is closely linked to another type of measurement called power. It is a convenient way to quantify the speed at which work is
https://fr.wikihow.com/calculer-le-travail-d%27une-force

Plane pulls are a thing, often for charity, so I wouldn't say they've never been done by men, though those are usually pulling the plane forward with the use of a rope and not back.

Yeah maybe you're right

Could you put some numbers in these equations to come up with an answer for how many people it might take?

In this situation acceration seems less of an issue than simply overcoming static friction and wheel rolling resistance.  Accepting a very very low acceleration rate would still not allow one single person to accomplish the task unaided!

@ZachLipton there's nothing that says that a "Plane pull" has to be forward in relation to the plane. Assuming there's enough room under the belly for the puller(s), the "pull" could be backwards in relation to the direction the plane is pointing. Ergo, a "plane pull" is entirely valid for this example. Also, 2 of these super freaks could each have a rope, one standing on either side of the fuselage, and, working together, neither should have to break a sweat!


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Is the A-10's cannon effective against tanks?
Perhaps this isn't on-topic, but is the A-10 effective against tanks (tank armour) when using its cannon?
For example, this page claims:

Using the cannon, the A-10 is capable of disabling a main battle tank from a range of over 6,500 m.

Is that true?
If that were so, why does a "main battle tank" (e.g. an M1 Abrams) have a 120 mm gun -- contrast that with the A-10's 30 mm cannon?
I thought that in WW2 you needed something like a 75 mm or 88 mm gun to be effective against the then-heavy tank armour. And the gun on a "main battle tank" has if anything increased in size since then -- or converted to missile instead of projectile -- is a 30 mm cannon effective?
Is the armour-penetration of an A-10 exaggerated (if it is effective against only e.g. softer armour than main battle tanks)? Does the cannon being mounted in a plane make it especially effective (I'd guess not, the muzzle velocity of a GAU-8 is 1000 m/s and and A-10's airspeed only adds 200 m/s to that)? If it is effective then why isn't that the primary armament of a main battle tank?

This question is as much about, "why aren't tanks equipped with gigantic 30mm gatling guns" as anything - which is way, way off topic. That the A-10 is effective as a tank killer is, imho, sufficiently on-topic, but the explanations for why tanks don't use the guns that aircraft do, and vice versa, really isn't.

I think it's more-or-less on-topic, e.g. Why do fighter jets still have guns/cannons? seems purely about armament rather than flying.

The last question was an afterthought to express scepticism -- "If a 30 mm cannon were really effective against main battle tanks, then wouldn't they be used by tanks, instead of the 120 mm guns that they actually use?"

Your source quoting 6500m for the gun seems wildly optimistic. Killing something at that range with a Maverick, possibly. But the wikipedia page says 4000m effective, 12,000m max range. To address the afterthought question, tanks handle one big recoil well; aircraft handle sustained recoil well. Neither would do well the other way around.

@RalphJ Your source quoting 6500m for the gun seems wildly optimistic Yes, I think so. "My source" was simply one of the top Google search results on the subject. And I'm even sceptical whether they're effective at any range against a "main battle tank" -- that's my question -- maybe something quite a bit older, or lightly-armoured personnel carrier -- that I could believe easily.

@RalphJ I thought you hit the nall on the head when you commented, "gigantic 30mm gatling guns" -- but per Wikipedia a GAU 8 has a barrel length of (only) 90.5 in (2.30 m), whereas an M1 has a Rheinmetall Rh-120 which is 5.3 or 6.6 m (i.e. 2 or 3 times as long).

Possible related questions- how effective were unguided rockets against tanks in WW2?  How effective were aircraft-mounted machine guns or  cannons against tanks in WW2?

@quietflyer Have those questions been asked and answered somewhere? I guess you're only saying "They were used, therefore they must have been effective."

I read a magazine article in last year or two claiming unguided rockets as carried on Typhoon etc were much less effective (due to missing targets) than generally believed- included interview w/ German tank soldier- don't recall source, maybe Aviation History.

@quietflyer Even as a schoolboy I was surprised that WW2 aircraft rockets were accurate enough to hit a vehicle-sized target (if they were) -- it didn't seem likely when they're unguided, I guessed less accurate than cannon. This Google search suggests they weren't accurate, that all airforces claimed more tank kills than ever actually happened, and that the chance of hitting a tank with a rocket was 5% in ideal conditions, or more like 1 in 800 in battle conditions.

Well it depends....
It depends on the tank, where you hit it, what angle you hit it, how far the A-10 was from the tank when it fired, and what kind of ammunition was used.
Wikipedia quotes that a GAU-8/A firing the PGU-14/B depleted uranium API ammunition offers the following armor penetration capability.

Armor penetration of Armor-Piercing Incendiary ammunition, BHN-300 RHA, attack angle 30 degrees from vertical:

76 mm at 300 meters
69 mm at 600 m
64 mm at 800 m
59 mm at 1,000 m
55 mm at 1,220 m

Whether that's a declassified spec with all other known values a secret or a realistic representation is unknown. I'd say it sounds reasonable. Tanks tend to have heavy armor on the front and sides but lighter on the top. Certainly the PGU-14/B could defeat armor on APCs and older generation MBTs. Modern MBTs with advanced composite/reactive armor may be more difficult to crack. While it might not be able to penetrate the chobham hull of an M1A2, it can certainly take out fuel tanks, destroy the engine, etc.
Keep in mind that a GAU-8/A can hammer a 5-mil Circle with 70 such rounds with a 1 second burst, provided the pilot has a steady trigger hand. This also can have a cumulative destructive effect on armor plating.
While the gun can be employed against armor, the preferred method is to attack tanks using standoff weapons like the AGM-65 Maverick missile. Getting close enough to tank columns to use the gun can be hazardous, especially if the group has mobile AAA/SAM escorts.

Thank you -- I hadn't noticed that in Wikipedia. What's a "5-mil" circle? The article you quoted says that the precision is such that "80% of rounds fired at 4,000 feet (1,200 m) range hit within a 40-foot (12 m) diameter circle". That's about 1 round per 2 square metres per second, unlikely to hit the same bit of armour twice. Also I think of main tank armour as being, like, 350 mm; and from videos I thought it would be not nearly as vertical as "30 degrees from vertical". I guess that answers my question though, thanks again.

I see -- "5-mil" is presumably a (small) unit of angle, meant to measure the scatter.

@ChrisW consider that splitting the armor and blowing up/killing the crew isn't the only method for disabling a tank. Destroying a tread, jamming the turret rotation mechanism or gun elevation mechanism will effectively render the tank useless until repaired. The crew will not want to be outside the tank fixing a tread while there are a pack of A-10s circling, especially if those A-10s are providing cover for nearby ground troops.

Yes, Wikipedia's articles say that the A-10 killed 900 tanks during the Gulf War, and that the Maverick was used 5000 times against armoured targets -- that's compatible with the answer's suggesting that the Maverick is the principle anti-tank weapon. The A-10 was designed in the 1970s, I'm guessing the cannon might be effective against any target possibly except a "main battle tank".

It effective against certain kinds of main battle tanks - whole lotta Iraqi T-72s got cut to ribbons with that thing.  But times, tank designs and armor changes.  Then again so do ammunition types, etc.

"While it might not be able to penetrate the chobham hull of an M1A2, it can certainly take out fuel tanks, destroy the engine, etc." - don't most modern tanks have the fuel tanks, engine, etc., inside the armoured hull, for precisely this reason?

@ChrisW A "mil" is a milliradian, of which there are approximately 6,283 in a circle.  1° ≈ 17.5 mil.  1 minute of angle ("moa") ≈ 3.44 mil.

As for the main battle tank's armour: it definitely is not 350mm. That is more than one foot of steel...

@Jpe61 it depends on the area of the tank we’re talking about.

Well that is technically true if we take into account the angle of impact, but it's still not the same.

@Jpe61 The angle of impact as well as the section of the tank that we’re talking about. You can’t apply armor plating the same thickness all over the tank as it would probably be too heavy to effectively maneuver in combat, so the armor protection has to be optimized for areas on the tank. Most tanks I believe were built for primary protection against other tank primary weapons.  Undoubtedly improvements to armor have been made to counteract airborne and missile threats with pop-up maneuver capabilities.

The thinnest armour on most MBTs is the roof of the turret and hull.  The A-10, when it attacks with its cannon, is diving from a fairly high aspect and therefore shooting at the most vulnerable parts, the roof and engine compartment.
Also, the rate of fire of the gun is so high and the projectiles are therefore so close together, there is a high probability of multiple hits on the same point. The result is an 30mm depleted uranium AP round, even if it doesn't penetrate the first time, has a good chance of being followed by another right behind it into the divot made by the first one and finishing the job.  An old GE video of the GAU8 on a test range shows the effect of follow up rounds hitting the crater made by their predecessors, and basically chewing their way through quite thick armour.

I haven't seen that video -- I've only ever seen videos of obsolete hulks of tanks being shot at, with debris and so on flying around, but without details of the damage. But perhaps you're right, various modern anti-armour munitions try to use Top attack. Effective against a "main battle tank" seems to me to be quite a tall claim though.

This is the correct answer. Tanks don't have more than 30-40 mm of armor on top. This armor isn't anywhere as sophisticated as frontal protection, and is either just steel or steel backed by a bit of lightweight composite material (fiberglass, aramid). 30mm AP rounds can penetrate MBT top armor.

Given the threat posed by air attack, why don't modern tanks have thick armour on the top as well as the sides?

@Sean Weight - something has to give somewhere. Modern warfare is generally a combined arms affair. Tanks have air support to deal with things like that.

@T.J.L.: So lighten it in some other way, like having it carry less ammo (a higher ammo capacity won't help you if it means you have to skimp on the armour and get blown up as a result), or simply give it a more powerful engine.

@Sean More powerful engines are heavier, too. As is the suspension to support all that extra weight. It's a balancing act. Rest assured that the military industrial complex has taken all of these factors into consideration when designing combat equipment... within the government's requirements and budgets, anyway.

Define effective

Using the cannon, the A-10 is capable of disabling a main battle tank from a range of over 6,500 m.

While modern battle tanks have quite sophisticated armor protection (for the crew), there are many vital components with less protection. In the image below, we see the tracks and the wheels, which are exposed. We see some optical systems for awareness and fire control.
If an aircraft manages to damage enough of these exposed systems to render the tank incapable of fulfilling its role, then the tank has been effectively knocked out, despite it being not killed.
The quote of the OP talks of disabling, not destroying. Thus, making the tank blind, or unable to move, may be sufficient to disabling it.

Image source: Wikimedia Commons

First of all adding "only" 200 m/s aircraft speed to 1000 m/s barrel velocity will give 40% more kinetic energy to those depleted uranium rounds.
Secondly, an aircraft can attack at many more angles than another tank, more easily finding thinner armor points.
Thirdly, even if aircraft could carry an 88mm gun, it has a much higher probability of a miss as it is moving in three dimensions.  There for, it fires a guided missile at longer range and closes to point blank with its Gatling, accomplishing its task with multiple hits.
A tank on the ground can aim much better with its gun and goes for one good shot.
But nothing wrong with helicopters carrying missiles to help out, along with the Warthogs.
As long as the A-10 is effective, it will still be around.

good math but wrong logic
not 40% but 44% 1000x 1000 =1000000
1200x1200= 1440000 that is 44% increase You did 200x200= 400000 that is your 40% increase (wrong logic)